Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11470 Accepted Submission(s): 7136
Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can‘t move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
‘.‘ - a black tile
‘#‘ - a red tile
‘@‘ - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
Asia 2004, Ehime (Japan), Japan Domestic
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ac代码
#include<stdio.h>
#include<string.h>
char map[22][22];
int n,m,ans;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
void dfs(int x,int y)
{
int i;
ans++;
map[x][y]='#';
for(i=0;i<4;i++)
{
int tx=x+dx[i];
int ty=y+dy[i];
if(tx>=0&&tx<n&&ty>=0&&ty<m&&map[tx][ty]=='.')
{
dfs(tx,ty);
}
}
}
int main()
{
while(scanf("%d%d",&m,&n)!=EOF,n||m)
{
int i,j,x,y;
for(i=0;i<n;i++)
{
scanf("%s",map[i]);
for(j=0;j<m;j++)
{
if(map[i][j]=='@')
{
x=i;
y=j;
}
}
}
ans=0;
dfs(x,y);
printf("%d\n",ans);
}
}